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551. A container (open at the top) made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find
(i) the cost of milk when it is completely filled with milk at the rate of Rs. 15 per litre.
(ii) the cost of metal sheet used, if it costs Rs. 5 per 100 cm². (Take π = 3.14)

Let l be the slant height of the frustum. We have r = 8 cm,
R = 20 cm and h = 16 cm

$therefore space space space l italic space space equals space space square root of left parenthesis straight R minus straight r right parenthesis squared plus straight h squared end root space space space space space space space space equals space space square root of left parenthesis 20 minus 8 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root space space space space space space space space equals space space space square root of left parenthesis 12 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root equals square root of 144 plus 256 end root space space space space space space space space equals square root of 400 space equals space 20 space cm. space space$

Now, volume of the container

$equals straight pi over 3 left parenthesis straight r squared plus straight R squared plus straight R. straight r right parenthesis space straight x space straight h equals space straight pi over 3 space left curly bracket 8 squared plus 20 squared plus 20 space straight x space 8 right curly bracket space straight x space 16 space cm cubed equals space fraction numerator 3.14 space straight x space 624 space straight x space 16 over denominator 3 end fraction space cm cubed$
$rightwards double arrow$ Volume of container = 10449.92 cm³

$rightwards double arrow$ Volume of container = $fraction numerator 10449.92 over denominator 1000 end fraction space litres$

$rightwards double arrow$ Volume of container = 10.44992 litres
= 10.45 litres (Approx)

∴ Cost of milk at the rate of Rs. 15 per litre = Rs.
(10.45 x 15) = Rs. 156.75
Now, Total surface area of the frustum = π (R + r) l + $straight pi$ r²= {3.14 (20 + 8) x 20 + 3.14 x 8²} cm²= 3.14 x (560 + 64) cm³= (3.14 x 624) cm³ = 1959.36 cm² $therefore$ Cost of metal used = Rs. $open parentheses fraction numerator 1959.36 space straight x space 5 over denominator 100 end fraction close parentheses$
= Rs. 97.96 (Approx)

1958 Views

552. An open container made up of a metal sheet is in the form of a frustum of a cone of height 8 cm with radii of its lower and upper ends as 4 cm and 10 cm respectively.
Find the cost of oil which can completely fill the container at the rate of Rs. 50 per litre. Also, find the cost of metal used, if it costs Rs. 5 per 100 cm² (Use π = 3.14).

1376 Views

553.

A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the volume of the bucket. Also find the cost of the bucket if the cost of metal sheet used is Rs. 20 per 100 cm². (Use $straight pi$ = 3.14)

6202 Views

Short Answer Type

554. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two places of decimals. Also find the total surface area of the remaining solid.
(Take π = 3.1416).

2086 Views

Long Answer Type

555.

A cylindrical vessel, with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of
(i) Water displaced out of the cylinderical vessel,
(ii) Water left in the cylindrical vessel.

$open square brackets Use space straight pi space equals space 22 over 7 close square brackets$

86 Views

556. A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm³. The radii of the top and bottom circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of metal sheet used in making it. (Use

straight pi

= 3.14)

4737 Views

Short Answer Type

557. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. If the total height of the toy is 15.5 cm. Find the total surface area.

open parentheses bold Use bold space bold pi bold space bold equals bold space bold 22 over bold 7 close parentheses

106 Views

558. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52 m and slant height of the conical portion is 53 m, find the area of the canvas needed to make the tent.

open parentheses bold Use bold space bold pi bold space bold equals bold space bold 22 over bold 7 close parentheses

70 Views

559. A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 25 cm. Find the total surface area of the vessel.

open parentheses bold Use bold space bold pi bold space bold equals bold space bold 22 over bold 7 close parentheses

131 Views

560. An iron pole consisting of a cylindrical portions 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the total surface area of the pole.

open parentheses bold Use bold space bold pi bold equals bold 22 over bold 7 close parentheses

99 Views